Answer:
Probability that no more than 2 out of 12 parts tested are defective is 0.8891.
Step-by-step explanation:
We are given that the probability that a machine part is defective is 0.1.
Twelve parts are selected at random.
The above situation can be represented through binomial distribution;
![P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......](https://tex.z-dn.net/?f=P%28X%20%3D%20r%29%20%3D%20%5Cbinom%7Bn%7D%7Br%7D%20%5Ctimes%20p%5E%7Br%7D%20%5Ctimes%20%281-p%29%5E%7Bn-r%7D%3Bx%3D0%2C1%2C2%2C3%2C.......)
where, n = number trials (samples) taken = 12 parts
r = number of success = no more than 2
p = probability of success which in our question is probability that
a machine part is defective, i.e; p = 0.1
<u><em>Let X = Number of machine parts that are defective</em></u>
So, X ~ Binom(n = 12, p = 0.1)
Now, Probability that no more than 2 out of 12 parts tested are defective is given by = P(X
2)
P(X
2) = P(X = 0) + P(X = 1) + P(X = 2)
= ![\binom{12}{0} \times 0.1^{0} \times (1-0.1)^{12-0}+\binom{12}{1} \times 0.1^{1} \times (1-0.1)^{12-1}+\binom{12}{2} \times 0.1^{2} \times (1-0.1)^{12-2}](https://tex.z-dn.net/?f=%5Cbinom%7B12%7D%7B0%7D%20%5Ctimes%200.1%5E%7B0%7D%20%5Ctimes%20%281-0.1%29%5E%7B12-0%7D%2B%5Cbinom%7B12%7D%7B1%7D%20%5Ctimes%200.1%5E%7B1%7D%20%5Ctimes%20%281-0.1%29%5E%7B12-1%7D%2B%5Cbinom%7B12%7D%7B2%7D%20%5Ctimes%200.1%5E%7B2%7D%20%5Ctimes%20%281-0.1%29%5E%7B12-2%7D)
= ![1\times 1 \times 0.9^{12}+ 12 \times 0.1^{1} \times 0.9^{11}+ 66\times 0.1^{2} \times 0.9^{10}](https://tex.z-dn.net/?f=1%5Ctimes%201%20%5Ctimes%200.9%5E%7B12%7D%2B%2012%20%5Ctimes%200.1%5E%7B1%7D%20%20%5Ctimes%200.9%5E%7B11%7D%2B%2066%5Ctimes%200.1%5E%7B2%7D%20%20%5Ctimes%200.9%5E%7B10%7D)
= 0.8891
Therefore, probability that no more than 2 out of 12 parts tested are defective is 0.8891.